MAT2400 Assignment 2 - Solutions
Notation: For any function f of one real variable, f (a+ ) denotes the limit of f (x)
when x tends to a from above (if it exists); i.e., f (a+ ) = limt→a+ f (t). Similarly, f (a− )
denotes the limit of f (x) when x tends to a from below (if it exists).
Problem 1.
The aim of this problem is to study a phenomenon which is called Gibb’s phenomenon.
At every simple jump discontinuity of a function f , the partial sums of the Fourier
series of f “overshoots” near the singularity by an amount about 9% of the “jump” of
the function.
To be presise, assume that f (x) has a jump singularity at a; i.e., d = f (a+ ) −
f (a− ) �= 0 and is continuous elsewhere in a neighbourhood of a. For simplicity we
assume that d > 0. We let sn (x) be the n-th partial sum of the Fourier series of f .
Then there is a sequence {xn } tending to a from above such that sn (xn ) > f (a+ ) + αd,
where the constant α satisfies α ≈ 0.089, i.e., about 9%. There is a similar sequence
{yn } tending to a from below with sn (yn ) < f (a− ) − αd
In this this problem we will study Gibbs phenomenon for the particular function
given in (−π, π) by:


if 0 < x < π
π/2
d(x) = 0
if x = 0


−π/2 if − π < x < 0
a) Compute the Fourier coefficients of d, and show that we have the equality
d(x) = 2
∞
�
sin (2k − 1)x
k=1
for all x ∈ (−π, π).
2k − 1
Assignment 2 — solutions— solutions
MAT2400 — spring 2012
Solution: The function is odd, so its Fourier series is a pure sine-series, and we need
only compute
�π
�
�
� 1
1 π
2 ππ
(1 − (−1)n )
bn =
d(x) sin nx dx =
sin nx dx = �� (− cos nx) =
,
π −π
π 0 2
n
0n
which equals 0 if n is even and
is
2
n
if n is odd. This gives that the Fourier series of d(x)
2
∞
�
sin (2k − 1)x
2k − 1
k=1
.
Clearly the function d(x) has one-sided derivatives everywhere, hence by Dini’s test
(or one of the corollaries, Corollary 14.12.4 in Tom’s) the Fourier series converges to
(d(x+ ) + d(x− ))/2 for every x, but this equals d(x) for all x.
b) Let the partial sums of the Fourier series of d(x) be denoted by dn (x). Show that
we have
� x
n
�
sin (2k − 1)x
sin 2nt
dn (x) = 2
=
dt.
2k
−
1
sin
t
0
k=1
�
Hint: Compute the derivative of dn (x) and use that 2 nk=1 cos(2k − 1)x = sinsin2nx
. To
x
prove the last formula, use the for us now well used and classical formula 2 sin α cos β =
sin(β + α) − sin(β − α).
Solution:
n
�
n
1 �
sin 2nx
2
cos(2k − 1)x =
(sin 2kx − sin 2(k − 1)x) =
sin x k=1
sin x
k=1
for x �= 0, π or −π (use the formula in the hint repeatedly with β = (2k − 1)x and
α = x), and, in fact, if we interpret the right side as the appropriate limit lim sinsin2nx
, it
x
holds as well for x = ±π (both sides are zero) and for 0 (both sides are 2n). Computing
the derivative of dn (x) term by term, we get
d�n (x)
=2
n
�
k=1
and integrating, we obtain
dn (x) =
�
cos(2k − 1)x,
0
x
sin 2nx
.
sin x
—2—
Assignment 2 — solutions— solutions
MAT2400 — spring 2012
c) Show that for t ≥ 0 the following inequality holds true
0 ≤ t − sin t ≤ t3 /6.
Use that inequality to prove that
�
�
� 1
�
1
�
� ≤ π t,
−
� sin t t � 12
when 0 < t ≤ π/2.
Solution: It is classical that sin t ≤ t for all t ≥ 0. To show the other inequality we
let
f (x) = t − sin t − t3 /3!
and compute f � (t) = 1 − cos t − t2 /2 and f �� (t) = sin t − t which is negative for t > 0.
Hence f � (t) < 0 for t > 0 since f � (0) = 0. It follows that f (t) < 0 for t > 0 since
f (0) = 0. We know that 2t
≤ sin t for 0 ≤ tπ/2, so we get
π
�
� �
�
� 1
1 �� �� t − sin t ��
π
π
3
�
� sin t − t � = � t sin t � ≤ 2t2 · t /6 = 12 t.
d) Prove that for all n and all 0 < x < π/2:
�
�
�
�dn (x) −
�
0
2nx
�
sin u ��
π
du� < x2
u
24
and use this to prove that for a given � > 0 there is an n0 such that if n ≥ n0 , then
dn (π/2n) > π/2 + απ − �
�π
where the constant α is given by α = π −1 ( 0
sin u
u
du − π/2). Hence
dn (π/2n) ≥ π/2 + 0.089π.
because one may compute α = 0.08949 . . . . (You can consider that value as given!).
—3—
Assignment 2 — solutions— solutions
MAT2400 — spring 2012
Solution: Integrating the inequality in d), we get
�� x
� � x
� x
�
�
sin
2nt
sin
2nt
πt
π
�
dt −
dt�� ≤
= x2 .
�
sin t
t
24
0
0
0 12
Substituting u = 2nt in the second integral and using xxx, we get
�
�
� 2nx
�
π
sin u ��
�dn (x) −
du� ≤ x2 .
�
u
24
0
Now, we put x = π/2n in the formula above to get
� π
sin u
π 3 −2
dn (π/2n) ≥
du − n > π/2 + απ − �
u
96
0
once n is so big that
π 3 −2
n
96
< �.
Problem 2.
Let C = C([0, 1], R) be the Banach space of continuous real valued functions on the
interval [0, 1] with norm given by �f � = sup{|f (x)| : x ∈ [0, 1]}. Fix an element g ∈ C,
and let I : C → C be the map given by
I(f )(x) =
�
x
f (t)g(t) dt.
0
a) Show that I is a bounded linear map; that is, I is linear and there is a positive
constant M such that �I(f )� ≤ M �f � for all f ∈ C. Determine the least such constant
if g is a positive function.
—4—
Assignment 2 — solutions— solutions
MAT2400 — spring 2012
Solution: I is linear by wellknown properties of the integral (in fact linearity). To
see that I is bounded, we compute
� � x
�� x
�
�
|f (t)g(t)| dt ≤
f (t)g(t) dt�� ≤
|I(f )(x)| = ��
0
0
� 1
≤
|f (t)g(t)| dt ≤ sup |f (t)g(t)|
0
≤ sup |f (t)| sup |g(t)| = �f � M,
where M = sup |g(t)|. Hence �I(f )� ≤ �f � M . If the function g is positive, we compute
�I(1)� = sup |g(t)|. Hence M = sup |g(t)| is the smallest constant we can use.
b) Show that the map I : C → C is uniformly continuous.
Solution: Let � > 0 be given, and let the corresponding δ > 0 be δ = �/M . Then
�I(f ) − I(g)� ≤ M �f − g� < M · �/M = �
whenever �f − g� < δ,
c) Show that for any bounded subset A ⊆ C the set I(A) ⊆ C is equicontinuous.
Solution: Let K be a bound for A, that is �f � ≤ K for all f ∈ A. We have
�� x
� � x
�
�
�
|I(f )(x) − I(f )(y)| = �
f (t)g(t) dt�� ≤
|f (t)g(t)| dt ≤ |x − y| KM
y
y
for f ∈ A. Then, given � > 0, we put δ = �/KM , and obtain
|I(f )(x) − I(f )(y)| ≤ |x − y| KM ≤ �/KM · KM = �
once |x − y| < δ, and this holds for all f ∈ A.
—5—
Assignment 2 — solutions— solutions
MAT2400 — spring 2012
d) Show that the closure I(A) is a compact subset.
Solution: We want to apply the Arzela-Ascoli theorem. Now I(A) is equicontinuous
since I(A) is; indeed, if � > 0 is given, choose δ > 0 such that |I(g)(x) − I(g)(y)| < �/3
for all g ∈ A and for all |x − y| < δ. Pick an element F ∈ I(A) and let I(fn ) be a
sequence converging (uniformly) to F . We have
|F (x) − F (y)| ≤ |F (x) − I(fn )(x)| + |I(fn )(x) − I(fn )(y)| + |I(fn )(y) − F (y)|
Let � > 0 be given. Choose N such that n > N gives |F (x) − I(fn )(x)| < �/3 for all x.
Then we get by the above inequality.
|F (x) − F (y)| < �.
Too see that I(A) is bounded, use that the norm is continuous, hence if I(fn ) converges
to F , then �F � = limn→∞ �fn � < KM . It follows from the A&A theorem, that I(A)
is compact. (It is closed by definition).
e) For each real number λ �= 0, let Vλ = {f ∈ C : I(f ) = λf }. Show that Vλ is a
subvector space of C. Determine all functions in Vλ .
Solution: It is clear that Vλ is a sub vector
� xspace (closed under addition and scalar
multiplication). An element f lies in Vλ if 0 f (t)g(t) dt = λf . The left side of this
equation is differentiable (integrals of continuous functions are) hence f is differentiable,
R
1 x
g(t) dt
λ 0
and λf � = f g. This is a first order
differential
equation
with
solution
f
(x)
=
Ce
�x
if λ �= 0, but since λf (x) = 0 f (t)g(t) dt, we see that f (0) = 0, hence C = 0, and
f ≡ 0; meaning that Vλ = 0. If λ = 0, it is a little more complicated. Then we get
f (x)g(x) ≡ 0, hence V0 is the subspace {f : f (x)g(x) ≡ 0}; and if e.g., g is positive,
we get f ≡ 0.
Problem 3.
Let F (x) be a strictly increasing function. For any half open interval I = (a, b ] define
m(I) = F (b) − F (a), and for any set E ⊆ R, let
�
ν ∗ (E) = inf{
m(I) : A}
I∈A
where A runs through all countable coverings of E by half open intervals (a, b ].
a) Show that ν ∗ (E) ≥ 0, and that ν ∗ is monotone; i.e., ν ∗ (E � ) ≤ ν ∗ (E) whenever
E � ⊆ E.
—6—
Assignment 2 — solutions— solutions
MAT2400 — spring 2012
Solution: Since F is increasing, m(I) = F (b) − F (a) > 0. Hence ν ∗ (E) ≥ 0, ν ∗ (E)
being the supremum of a set of positive numbers. If E � ⊆ E, then any covering of E
(of the type we use) is also a covering of E � (of the type we use). Hence ν ∗ (E � ) is the
supremum of a smaller set than ν ∗ (E), so ν ∗ (E � ) ≤ ν ∗ (E).
b) Show that ν ∗ is semiadditive; that is
∗
ν (
∞
�
n=1
En ) ≤
∞
�
ν ∗ (En )
n=1
for any family {En } of subsets of R.
Solution: This is word by word the same proof as of Proposition 5.1.4 page 146
in Tom’s notes. Take a look at that.
c) If x ∈ R, show that ν ∗ ({x}) = F (x) − F (x− ), and hence ν ∗ {x} = 0 if and only if F
is continuous from the left at x.
Solution: The sequence F (x − 1/n), where n ∈ N, is increasing with F (x− ) as limit,
hence F (x − 1/n) ≤ F (x− ) for all n. Any half open interval (a, b] containing x contains
an interval of the form (x − 1/n, x] where n ∈ N. Hence
m(I) = F (b) − F (a) ≥ F (x) − F (x − 1/n) ≥ F (x) − F (x− )
This shows that ν ∗ ({x}) ≥ F (x) − F (x− ). On the other hand, ν ∗ ({x}) ≤
m((x − 1/n, x]) = F (x) − F (x − 1/n) for all n, hence ν ∗ ({x}) ≤ inf n∈N {F (x) − F (x −
1/n)} = F (x) − F (x− ); and thus ν ∗ ({x}) = F (x) − F (x− ). The function F is continuous from the left at x if and only if F (x− ) = F (x) , hence if and only if ν ∗ {x} = 0 ,
by what we just saw.
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